pub fn new_birthday_probability(n: u32) -> f64 {
    const DAYS_IN_YEAR: u32 = 365;
    // 首先算出来所有人都不在同一天的整体概率
    let no_match_prob = (0..n)
        .map(|i| (DAYS_IN_YEAR - i) as f64 / DAYS_IN_YEAR as f64)
        .product::<f64>();
    // 然后取反，就是至少有两个人在同一天过生日的概率
    1.0 - no_match_prob
}
